heureka

$$\\ \left[ 5-{3\times12\over(5\times5+15)} +(3\times0+40) \right] \times(5\times5 \times{3\over15} )\\\\ =(5-{36\over40}+40})\times(5\times {15\over15})\\\\ =(45-{9\over10}})\times5\\\\ =45\times5-{45\over10}}\\\\ =225-4.5\\\\ =220.5$$

how to convert 486mm to m:

A parabola is defined by the equation . In which direction will the parabola open?

$$\boxed{y=ax^2+bx+c} \qquad \mbox{ parabola}$$

a > 0     opens upward 

a < 0     opens downward

$$\\x^2={3 \over 4}y\\\\ y=\underbrace{{4 \over 3}}_{ a > 0} x^2$$

a > 0 parabola opens upward

e^2x + 1 = 900   ?

$$\begin{array}{rcrc} e^{2x+1} &=& 900 & \quad | \quad \ln{}\\ \ln{(e^{2x+1} )}&=&\ln{( 900 )} & \\ (2x+1)*\ln{( e )}&=&\ln{( 900 )} & \quad |\quad \ln{(e)} = 1 \quad !\\ 2x+1 &=&\ln{( 900 )} & \\ 2x&=&\ln{( 900 )} - 1& \\ x&=&{ \ln{(900)} -1 \over 2 } &\\ x&=&2.90119738166&\end{array}$$

9^−x/200 = 2  ?

$$\begin{array}{rcrc} 9^{ (-{x\over 200}) } &=& 2 & \quad | \quad \ln{}\\ \ln{( 9^{ (-{x\over 200}) } ) }&=&\ln{(2 )} & \\ (-{x\over 200}) } *\ln{(9 )}&=&\ln{(2 )} & | \quad *\quad -({200\over \ln{(9)}})\\ x&=&-200*{\ln{(2)}\over\ln{(9)}} &\\ x&=&-63.0929753571& \end{array}$$

7^0.9x = 6   ?

$$\begin{array}{rcrc} 7^{0.9x} &=& 6 & \quad | \quad \ln{}\\ \ln{( 7^{0.9x} )} &=& \ln{( 6 )} & \\ 0.9*x*\ln{( 7 )} &=& \ln{( 6 )} & \\ x&=&{ \ln{(6)} \over 0.9 *\ln{(7)} } &\\ x&=&1.02309135685& \end{array}$$   

x ² 3 ^x −  4( 3 ^x) = 0    ?

$$\\x^2 3^x-4(3^x)=0 \\ 3^x ( x^2-4 ) =0 \\ 3^x ( x-2 ) ( x+2 ) =0\\\\ \underbrace{3^x}_{=0} \times( \underbrace{x-2}_{=0} ) \times( \underbrace{x+2}_{=0} ) =0 \\\\ \boxed{3^x=0} \quad | \quad \ln{} \\\\ \ln{(3^x)} = \ln{(0)} \\ x\ln{(3)} = \ln{(0)} \\ x= { \ln{(0)} \over \ln{(3)} } } \quad | \quad \ln{(0)} \mbox{ no solution !}\\\\ \boxed{x-2=0} \quad \Rightarrow \quad \boxed{x=x_1=2}\\\\ \boxed{x+2=0} \quad \Rightarrow \quad \boxed{x=x_2=-2}$$

e^6x + 5e^3x − 14 = 0 ?

$$\\e^{6x}=e^{3x}e^{3x} \quad | \quad \mbox{ set }\quad \boxed{z=e^{3x}}\\ z^2 +5z -14=0\\ \underbrace{1}_{a=1}z^2 \underbrace{+5}_{b=5}z \underbrace{-14}_{c=-14} =0\\ \boxed{z_{1,2}= { -b\pm\sqrt{b^2-4ac} \over 2a } }\\\\ z_{1,2}= { -5\pm\sqrt{25-4*1*(-14)} \over 2*1 } \\\\ z_{1,2}= { -5\pm\sqrt{25+56} \over 2} \\\\ z_{1,2}= { -5\pm9\over 2} \\\\ \boxed{z_1= 2 \quad z_2 = -7} \\\\ e^{3x}=z \quad | \quad \ln\\\\ \ln{(e^{3x})}=\ln{(z)}\\\\ 3x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\ \boxed{x={1\over 3}\ln{(z)}} \\\\$$

$$\\x=x_1={ \ln{(2)}\over 3} } =0.23104906019\\ x=x_2={ \ln{(-7)}\over 3} } = \mbox{no solution !}\\\\ \boxed{x=0.23104906019}$$

e^2x − 9e^x + 8 = 0  ?

  $$\\e^{2x}=e^xe^x \quad | \quad \mbox{ set }\quad \boxed{z=e^x}\\ z^2 -9z +8=0\\ \underbrace{1}_{a=1}z^2 \underbrace{-9}_{b=-9}z \underbrace{+8}_{c=8} =0\\ \boxed{z_{1,2}= { -b\pm\sqrt{b^2-4ac} \over 2a } }\\\\ z_{1,2}= { 9\pm\sqrt{81-4*1*8} \over 2*1 } \\\\ z_{1,2}= { 9\pm\sqrt{81-32} \over 2} \\\\ z_{1,2}= { 9\pm7\over 2} \\\\ \boxed{z_1= 8 \quad z_2 = 1} \\\\ e^x=z \quad | \quad \ln\\\\ \ln{(e^x)}=\ln{(z)}\\\\ x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\ \boxed{x=\ln{(z)}} \\\\$$

$$\\x=x_1=\ln{(8)}=2.07944154168\\ x=x_2=\ln{(1)}=0\\ \boxed{x_1=2.07944154168 \qquad x_2=0}$$

why is 1 not a prime number? why?

Prime numbers are never divisible by other prime numbers. If was 1 a prime number, would be valid this no more, because every number is divisible by 1